A Series of Hard Problems (3 of 3)

Number Theory Level 3

Suppose you have a 20 × 16 bar of chocolate squares. You want to break the bar into smaller chunks, so that after some sequence of breaks, no piece has an area of more than 5. What is the minimum possible number of times that you must break the bar?

63 45 21 65

1 solution

Vishruth Bharath
Jan 6, 2018

First, we will prove that it is not possible to use fewer than $63$ cuts. Notice that in order to satisfy the conditions, we will need at least $\frac{20*16}{5}=64$ pieces. Each cut will create exactly $1$ more piece. Therefore, we need at least $63$ cuts. In order to see that it can be done in $63$ cuts, first, make $15$ cuts, each one parallel with the side of length $20$ , so that we have $16$ strips of length $20$ . Cutting each of these strips into $4$ equal parts will give us only chunks of area $5$ , and will take $3$ cuts per chunk. We make $15+ 3(16) \rightarrow \boxed{63}$ cuts total in order to do this.

Problem is Courtesy of The BMT (Berkeley Math Tournament)

A Series of Hard Problems (3 of 3)

1 solution

0 pending reports